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  • OT Maths Question

    I have found a curve fit for some experiments that I have done and got the equation.

    Y = (A x ( Flow^B))+ (C x (Flow^D))

    Can anyone help in transposing the formulae to give Flow=, I cant remember maths to this level although I can still transpose trig and other simpler equations.

    Steve Larner

  • #2
    More Info.

    What method did you use to process the data to obtain the equation? I've never heard of the "Flow" function. I just might be something that is peculiar to the function/routine which you used to generate the equation.
    Bill

    Being ROAD KILL on the Information Super Highway and Electronically Challenged really SUCKS!!

    Every problem can be solved through the proper application of explosives, duct tape, teflon, WD-40, or any combo of the aforementioned items.

    Comment


    • #3
      Originally posted by BigBoy1
      What method did you use to process the data to obtain the equation? I've never heard of the "Flow" function. I just might be something that is peculiar to the function/routine which you used to generate the equation.
      Sorry Flow is the property that is x on the graph, should read

      Y = (A x ( X^B))+ (C x (X^D))

      Steve Larner

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      • #4
        Is little x and big X the same or is little x a multiplication operator?

        In other words, is it Y = AX^B + CX^D
        Last edited by dp; 04-15-2010, 12:34 PM.

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        • #5
          Originally posted by dp
          Is little x and big X the same or is little x a multiplication operator?

          In other words, is it Y = AX^B + CX^D
          little x is multiplication


          Y = (A *( X^B))+ (C * (X^D))

          Steve Larner

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          • #6
            would you be referring to the eqn of continuity [abridged] leading to an all out assault on Bernoulli's[spelling?], ie cross section at point A x velocity [or mass flow rate] = cross section at point B x velocity [or mass flow rate]
            AKA A1V1 =A2V2, it is reminiscent of a venturi meter
            mark

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            • #7
              Steve,

              It can be solved with logarithms. I got in my days practice at using maxima, the free computer arithmetic package coaxing it to do the math for me (twice as hard as actually doing the problem).

              Maxima says

              x=A^(1/(D-B))/((-C)^(1/(D-B))

              --Cameron

              Comment


              • #8
                The equation cannot in general be transposed to give a solution for X as a function of Y. Only if the exponents B and D are integers can solutions can be obtained, and then only if these integers are not greater than 4.

                This article states that a quartic is the highest order polynominal that can be solved in the general case: http://en.wikipedia.org/wiki/Quartic_equation
                Allan Ostling

                Phoenix, Arizona

                Comment


                • #9
                  Originally posted by ckelloug
                  Steve,

                  It can be solved with logarithms. I got in my days practice at using maxima, the free computer arithmetic package coaxing it to do the math for me (twice as hard as actually doing the problem).

                  Maxima says

                  x=A^(1/(D-B))/((-C)^(1/(D-B))

                  --Cameron
                  What happened to Y?

                  -Pete
                  I just like to make stuff.

                  Comment


                  • #10
                    Also, Wolfram-Alpha seems appropriate.

                    Comment


                    • #11
                      Y = A*X^B + C*X^D is best treated transcendental (literally meaning "transcending algebra"). I suggest this by the nature of the exponents and their arbitrary values of B and D. We can still use algebra, but encapsulate the exponential functions with natural logarithms first.

                      Rules:
                      1) ln(x*y) = ln(x) + ln(y)
                      2) ln(x/y) = ln(x) - ln(y)
                      3 ln(x^r) = r*ln(x)
                      4) e^ln(X) = X

                      Take the natural log (ln) of both sides:
                      ln(Y) = ln(A*X^B) +ln(C*X^D)

                      Apply rule #1:
                      ln(Y) = ln(A)+ln(X^B) + ln(C) + ln(X^D)

                      Apply rule #3:
                      ln(Y) = ln(A)+B*ln(X) + ln(C) + D*ln(X)

                      Now the algebra:
                      Subtract ln(A) and ln(C) from both sides (also put the ln(X) terms on the left):
                      B*ln(X) + D*ln(X) = ln(Y) - ln(A) - ln(C)

                      Factor out the ln(X) terms:
                      (B+D)ln(X) = ln(Y) - ln(A) - ln(C)

                      Divide both sides by B+D:
                      ln(X) = [ln(Y) - ln(A) - ln(C)] / (B+D)

                      Apply rule #3:
                      ln(X) = [ln(Y/A*C)] / (B+D)

                      Apply rule #4:
                      X = e^[ln(Y/A*C) / (B+D)]

                      Note: this solution is only valid where A,B,C,D, Y are all greater than 0.

                      I would really double check this math before I made any investments based on it.

                      Comment


                      • #12
                        Y = A*X^B + C*X^D

                        Take the natural log (ln) of both sides:
                        ln(Y) = ln(A*X^B) + ln(C*X^D)

                        Well, there's your problem. The ln(p + q) is not equal to ln(p) + ln(q).

                        ln(p) + ln(q) = ln (p*q) (as you pointed out)

                        I didn't bother to check but, since the first step in your derivation is wrong, I have to presume your final answer is also wrong.
                        Regards, Marv

                        Home Shop Freeware - Tools for People Who Build Things
                        http://www.myvirtualnetwork.com/mklotz

                        Location: LA, CA, USA

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                        • #13
                          Originally posted by aostling
                          The equation cannot in general be transposed to give a solution for X as a function of Y. Only if the exponents B and D are integers can solutions can be obtained, and then only if these integers are not greater than 4.

                          This article states that a quartic is the highest order polynominal that can be solved in the general case: http://en.wikipedia.org/wiki/Quartic_equation
                          ^^^^^
                          what he said

                          Comment


                          • #14
                            Hi Steve,

                            Sorry for the goof in the result I posted earlier.

                            You could probably solve it numerically by Newton Raphson iteration but depending on what it looks like, it might be easier to just draw a graph and estimate.

                            --Cameron

                            Comment


                            • #15
                              Check please

                              Cameron,

                              please check this out and see if I've got it right or wrong.

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